Question

Let d(3, -2), e(-3, 1) and f(4, - 3) be the midpoints of the sides bc, ca and ab respectively of ∆abc. Then, find the coordinates of the vertices a, b and



c.

Answer 1

$\bold{GIVEN,}$

ΔABC has sides

▶BC with midpoint d(3 , -2)

▶CA with midpoint e(-3 , 1)

▶AD with midpoint f(4 , -3)

Let the coordinates of vertices be:-

▶A$\bold{(x_1 , y_1)}$

▶B$\bold{(x_2 , y_2)}$

▶C$\bold{(x_3 , y_3)}$

By midpoint theorem we know :-

◼$\bold{For \: midpoints \: of \: BC}$

▶$x = \boxed{ \bold{\frac{x_2 + x_3}{2}}}$

$3 = \frac{x_2 + x_3}{2}$

3 × 2 = $x_2 + x_3$

6= $x_2 + x_3$-----(1)

▶$y = \boxed {\bold{\frac{y_2 + y_3}{2}}}$

$-2 = \frac{y_2 + y_3}{2}$

2 × -2 = $y_2 + y_3$

-4 = $y_2 + y_3$----(4)

◼$\bold{For \: midpoints \: of \: CA}$

▶$x = \boxed{ \bold{\frac{x_1 + x_3}{2}}}$

$-3 = \frac{x_1 + x_3}{2}$

-3 × 2 = $x_1 + x_3$

-6= $x_1 + x_3$-----(2)

▶$y =\boxed{ \bold{ \frac{y_1 + y_3}{2}}}$

$1 = \frac{y_1 + y_3}{2}$

2 × 1= $y_1 + y_3$

2= $y_1 + y_3$----(5)

◼$\bold{For \: midpoints \: of \: AB}$

▶$x = \boxed{ \bold{\frac{x_1 + x_2}{2}}}$

$4 = \frac{x_1 + x_2}{2}$

4 × 2= $x_1 + x_2$

8= $x_1 + x_2$-----(3)

▶$y =\boxed{ \bold{ \frac{y_1 + y_2}{2}}}$

$-3 = \frac{y_1 + y_2}{2}$

2 × -3= $y_1 + y_2$

-6= $y_1 + y_2$----(6)

$\bold{FOR \: x \:CoRDINATES}$

On adding eq (1) ,(2) and (3)

$x_2 + x_3 + x_1 + x_3 + x_1 + x_2 = 6 + (-6) + 8$

$2x_2 + 2x_3 + 2x_1 = 8$

$2(x_2 + x_3 + x_1 )= 8$

$x_2 + x_3 + x_1 = \frac{8}{2}$

$\bold{x_2 + x_3 + x_1 = 4}$

◼On placing the value of eq (1)6= $x_2 + x_3$

$6+ x_1 = 4$

$x_1 = 4- 6$

$x_1 = -2$

◼On placing the value of eq (2) -6= $x_1 + x_3$

$-6+ x_2 = 4$

$x_2 = 4 + 6$

$x_2= 10$

◼On placing the value of eq (3) 8= $x_1 + x_2$

$8+ x_3 = 4$

$x_3 = 4 - 8$

$x_3 = -4$

$\bold{FOR \: y \:CoRDINATES}$

On adding eq (4) ,(5) and (6)

$y_2 + y_3 + y_1 + y_3 + y_1 + y_2 = -4 + 2 + (-6)$

$2y_2 + 2y_3 + 2y_1 =- 8$

$2(x_2 + x_3 + x_1 )= -8$

$x_2 + x_3 + x_1 = \frac{-8}{2}$

$\bold{x_2 + x_3 + x_1 = - 4}$

◼On placing the value of eq (4)-4= $y_2 + y_3$

$-4 + y_1 = 4$

$y_1 = 4 + 4$

$y_1 = 8$

◼On placing the value of eq (5) 2= $y_1 + y_3$

$2 + y_2 = 4$

$y_2 = 4 - 2$

$y_2= 2$

◼On placing the value of eq (6) -6= $y_1 + y_2$

$-6 + x_3 = 4$

$y_3 = 4 - 6$

$y_3 = - 2$

So,

$\bold{CORDINATES \: OF \: A = (-2 ,8)}$

$\bold{CORDINATES \: OF \: B = (10 ,2)}$

$\bold{CORDINATES \: OF \: C = (-4 ,-2)}$

Answer 2

Answer:

Step-by-step explanation: